$h(a, b) = 2^{a}(1 + b)$ What is the partial derivative of $h$ with respect to $b$ ? Choose 1 answer: Choose 1 answer: (Choice A, Incorrect) Incorrect $\ln(2)2^a(1 + b)$ This is what we get if we take the partial derivative with respect to $a$, but we were told to differentiate with respect to $b$. (Choice B, Incorrect) Incorrect $\ln(2) 2^a$ This is what we get if we take the mixed partial derivative $\dfrac{\partial}{\partial a} \left[ \dfrac{\partial h}{\partial b} \right]$, but we were told to differentiate with respect to $b$. (Choice C, Incorrect) Incorrect $2^a( 1 + \ln(2) + b\ln(2))$ This is what we get if we treat both $a$ and $b$ as variables and use the product rule. We were told to differentiate with respect to $b$, so we can pretend $a$ is a constant. (Choice D, Checked, Correct) Correct (selected) $2^a$
Solution: Taking a partial derivative with respect to $b$ means treating $a$ like a constant, then taking a normal derivative. $\begin{aligned} \dfrac{\partial h}{\partial b} &= \dfrac{\partial}{\partial b} \left[ 2^a(1 + {b}) \right] \\ \\ &= \dfrac{\partial}{\partial b} \left[ 2^a \right] + \dfrac{\partial}{\partial b} \left[ 2^a{b} \right] \\ \\ &= 0 + 2^a \end{aligned}$ Note that $\dfrac{\partial}{\partial b} \left[ 2^a \right] = 0$ since we're treating $a$ as a constant. In conclusion, $\dfrac{\partial h}{\partial b} = 2^a$